Problem: The value of $\sqrt{111}$ lies between which two consecutive integers ? Integers that appear in order when counting, for example 2 and 3.
Solution: Consider the perfect squares near $111$ . [ What are perfect squares? Perfect squares are integers which can be obtained by squaring an integer. The first 13 perfect squares are: $ 1,4,9,16,25,36,49,64,81,100,121,144,169$ $100$ is the nearest perfect square less than $111$ $121$ is the nearest perfect square more than $111$ So, we know $100 < 111 < 121$ So, $\sqrt{100} < \sqrt{111} < \sqrt{121}$ So $\sqrt{111}$ is between $10$ and $11$.